Limiting Sums (2 Viewers)

civilKRobbo · May 14, 2017

Hey could someone help me with this question?
The third term of a geometric series is 54 and the sixth therm is 11 and 83/125. Evaluate a and r and its limiting sum.

Thanks

1729 · May 14, 2017

civilKRobbo said:
Hey could someone help me with this question?
The third term of a geometric series is 54 and the sixth therm is 11 and 83/125. Evaluate a and r and its limiting sum.

Thanks

What is the 83/125?

Drongoski · May 14, 2017

civilKRobbo said:
Hey could someone help me with this question?
The third term of a geometric series is 54 and the sixth therm is 11 and 83/125. Evaluate a and r and its limiting sum.

Thanks

When someone poses an incompletely-specified question it shows he/she does not fully understand the question. Often a key part of the question is omitted, he/she being unaware it is an important part of the whole question.

InteGrand · May 14, 2017

1729 said:
What is the 83/125?

Drongoski said:
When someone poses an incompletely-specified question it shows he/she does not fully understand the question. Often a key part of the question is omitted, he/she being unaware it is an important part of the whole question.

$\noindent Possibly, the original poster meant $11\tfrac{83}{125}$.$

Drongoski · May 14, 2017

InteGrand said:
$\noindent Possibly, the original poster meant $11\tfrac{83}{125}$.$

That makes sense. You are a great decipherer.
In that case:

$ar^2 = 54 $ and $ ar^5 = 11\frac {83}{125} = \frac {1458}{125}\\ \\ \therefore \frac {ar^5}{ar^2} = r^3 = 0.126 \implies r = 0.6 $ and $ a = 150 \implies S_{\infty} = \frac {a}{1-r} = \frac {150}{1-0.6} = 375$

InteGrand · May 14, 2017

civilKRobbo said:
Hey could someone help me with this question?
The third term of a geometric series is 54 and the sixth therm is 11 and 83/125. Evaluate a and r and its limiting sum.

Thanks

Drongoski said:
That makes sense. You are a great decipher.

$\noindent Assuming this was the case, the third term is $ar^{2} = 54$ and the sixth term is $ar^{5} = 11\tfrac{83}{125}$. This implies that $r^{3} = \frac{ar^{5}}{ar^{3}} = \frac{11\tfrac{83}{125}}{54} = \frac{27}{125}$. Taking cube roots, we have $\boxed{r = \frac{3}{5}}$. Now$

$\begin{align*}ar^{2} &= 54 \\ \Rightarrow a &= \frac{54}{r^{2}}\\ \Rightarrow \boxed{a = 150}. \quad \Big{(}\text{as } r= \frac{3}{5}\Big{)}\end{align*}$

$\noindent The limiting sum is thus $\frac{a}{1 - r} = \frac{150}{1 - \frac{3}{5}} = 375$.$

Limiting Sums (2 Viewers)

civilKRobbo

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1729

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Drongoski

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InteGrand

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Drongoski

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InteGrand

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